Silly question about trig past pi/2

[motornoob101]

I know all the well know angles from 0 to \pi/2. However, past pi/2 I am rather clueless and have no idea what angle corresponds to what value.

For example, in trying to figure out tan^{-1}(-1) How does one know it is 5pi/4 and not 7pi/8? I was debating between measuring pi/4 from the negative x-axis (which would give me 7pi/8) and measuring pi/4 from the y axis,(which would give me 5pi/4) but as you can see, I choose the wrong value. Is there a trick to it? Thanks.

 

[sutupidmath]

well

arctan(-1)=-arctan(1)=-\frac{\pi}{4} so it means that \frac{\pi}{4} lies on the fourth quadrant of the unit circle. So you can get that value by subtracting it from 2\pi that is

2\pi-\frac{\pi}{4}=\frac{8\pi-\pi}{4}=\frac{7\pi}{4} because -pi/4 is the angle counted clockwise direction.
YOu know that it does not lie on the 3rd quadrant because tan(x) there is positive. There are other tricks too.

 

[JohnDuck]

That's the "problem" with inverse trig functions--since, for example, there are an infinite number of solutions for y to the equation \tan{y} = x, in order for f(x) = \arctan{x} to be a well-defined function, you have to consider it as mapping to a particular interval of angles. Without qualification, arctan typically returns values between \frac{-\pi}{2} and \frac{\pi}{2}, but any interval of length \pi is valid as long as your usage is consistent. There are similar considerations for arcsin and arccos.

Edit: Because I'm not sure if I made it clear, let me say this: \arctan{x} is usually defined as the function from the real numbers to the interval (\frac{-\pi}{2} , \frac{\pi}{2}) such that \tan{(\arctan{x})} = x for any real x.

 

[motornoob101]

Oh I forget that arctan is an odd function. Thanks. But what if the question asks for a value for arctan(-1) within pi? I would have tried the 7pi/4 approach if I wasn't restricted to 0 to pi. Thanks.

 

[sutupidmath]

Oh I forget that arctan is an odd function. Thanks. But what if the question asks for a value for arctan(-1) within pi? I would have tried the 7pi/4 approach if I wasn't restricted to 0 to pi. Thanks.

If you have to find the value within an interval of a length \pi that doent mean that your interval must be [0,\pi] , moreover the common interval for this problem is chosen like JohnDuck stated [-\frac{\pi}{2},\frac{\pi}{2}].

This means that also <br /> \frac{7\pi}{2} is within that interval.

 

[JohnDuck]

One more thing arctan(-1) has no value that lies on the interval [0,\pi].

\frac{3\pi}{4}?

 

[sutupidmath]

\frac{3\pi}{4}?

Blahhhhhh! Yup.